Prove that if $X_n \xrightarrow{P} c$, then $E(X_n) \to c$ for $X_n$ uniformly bounded I have been trying to prove that for a random variable that is uniformly bounded, i.e. $|X_n - c| <M$ for all $n$, convergence in probability to $c$ implies that

$$E\left(X_n \right) \to c$$

as well. This is quite intuitive as all probability mass is bounded away from infinity in this case, but I am having a hard time formalizing this notion. Could you please give me some hints? Please keep it simple, though. Thank you.
 A: Start with 
$$
E|X_n-c|\le M\cdot P[|X_n-c|>\epsilon]+\epsilon\cdot P[|X_n-c|\le\epsilon]\le M\cdot P[|X_n-c|>\epsilon]+\epsilon.
$$
Because $X_n$ converges in probability to $c$, $\lim_n P[|X_n-c|>\epsilon]=0$ for each $\epsilon>0$. It follows from the  inequality displayed above that
$$
\limsup_nE|X_n-c|\le\epsilon,
$$
for each $\epsilon>0$.
A: 
A necessary and sufficient condition for $L^1$ convergence is $X_n\xrightarrow{P} X$ and the sequence $(X_n)$ is uniformly integrable. 

So, it remains to show that $X_n$ is uniformly integrable (or equivalently that $Y_n=X_n-c$ is uniformly integrable). But it is straightforward that $$E[|X_n-c|\mathbf 1_{|X_n-c|\ge M}]=0\le ε$$ for any $ε>0$. So $E[X_n-c]\to 0$.

Edit: Since $|X_n-c|$ is a non-negative random variable you have that $$E|X_n-c|=\int_{0}^{+\infty}P(|X_n-c|\ge x)dx=\int_{0}^{M}P(|X_n-c|\ge x)dx$$ where the second equality is implied by $|X_n-c|<M$. So you can bound the integration limits on the RHS at $M$ (away from infinity) and hence you can take limits and interchange the order of limit and integration (I will do it with Fatou and $\le$ but you can do it directly, since you have convergence and bounded region of integration. Fatou works also with unbounded limits.): \begin{align}\limsup_{n\to+\infty} E|X_n-c|&=\limsup_{n\to+\infty}\int_{0}^{M}P(|X_n-c|\ge x)dx\\& \le \int_{0}^{M}\limsup_{n\to+\infty}P(|X_n-c|\ge x)dx=0\end{align} because $\limsup P(|X_n-c|\ge x)=\lim P(|X_n-c|\ge x)=0$ due to convergence in probability.
